today we're going to be talking about how to use integration by parts to prove a reduction
formula and in this particular problem we've been asked to use integration by parts to
prove the reduction formula that I've written here and if you're not familiar with reduction
formulas basically all they are are formulas that tell you how to deal with a specific
form of an integral so for example this reduction formula tells you that if you're given exactly
the integral ln of x raised to a constant n here is a constant like if it's two or three
or four or something like that it's not a variable but if you're given the integral
of ln of x raised to a constant and your integral is in exactly that form you can use this formula
to give you a reduction of your original integral in other words this right hand side here is
going to be a simpler version of this integral and essentially the way that they create this
formula is with integration by parts so we've been asked to recreate that proof and show
the reduction formula reduction formulas are interesting because obviously you're not going
to memorize tons and tons of reduction formulas in the hope that you get an integral in exactly
this form so it's better to know methods for solving integrals like integration by parts
and u-substitution and partial fractions as opposed to memorizing reduction formulas but
once you identify a reduction formula it can be helpful to use over and over again so I've
also written the integration by parts formula here so we can use as a reference but essentially
what we're going to do is apply integration by parts to the left hand side and they want
us to come up with what's on the right hand side so if you'll remember for integration
by parts what we need to do is identify values for u and dv in this case u is going to be
the quantity ln of x raised to the n and dv is just going to be equal dx so what we'll
say is that u is equal to ln of x raised to the n and remember n is a constant and dv
we'll say is equal to dx now since we're dealing with integration by parts remember that once
you've identified u and dv you take the derivative of u to get du so the derivative of u is called
du and in this case du the way that we find it is by applying chain rule to ln of x raised
to the n so we have to use chain rule to take the derivative of this value here remember
that chain rule tells us to take the derivative of our outside function leaving the inside
function completely untouched and then multiply by the derivative of the inside function so
in other words our outside function is basically something raised to the n power our inside
function is ln of x so if we had for example if our function instead was just x to the
n right the derivative of that would be n times x to the n minus one we would just use
power rule so essentially we're ignoring what's inside here and just taking the derivative
of the outside so we bring our exponent down in front here using power rule and we say
n times ln of x to the n minus one that's the derivative of the outside function we've
used power rule to take the derivative of the outside we've ignored for now the inside
ln of x but because we're using chain rule we have to multiply this result here by the
derivative of the inside function well ln of x is the inside function the derivative
of ln of x is one dived by x so we have to multiply by one over x and of course because
we're taking the derivative here we have to add dx so let's go ahead and erase this to
give ourselves a little bit more room so we have to add dx to this dv keep in mind was
just dx alright so that's our value for du we have to also take the integral of dv to
get v so we can use it in our formula and the integral of dx is simply x so now that
we've identified values for u and du and v and dv we can plug them into our formula so
we identified u and dv in our original integral and now we're going to plug in u v v and du
to the right hand side of our integration by parts formula here so we're going to take
u times v so we have u and v here and when we multiply ln of x to the n times x we'll
just get x times ln of x to the n so that's those two multiplied together then our formula
tells us to subtract the integral so subtract integral of v times du so we have v here which
is just x times du which n times ln of x to the n minus one times one over x dx we've
now plugged in all of the components of our integration by parts formula and keep in mind
that we need to simplify this so that it's in exactly the same form we see over here
on the right hand side of our reduction formula so just a couple things we need to do the
first of which is to realize that we can cancel and x from the numerator and denominator of
our integral here so these two go away and we also need to recognize that we have n here
which remember is a constant n is not a variable it's a constant and because it's a constant
coefficient it can be moved out here in front of the integral so what we're left with is
x times ln of x raised to the n minus we move the n out in front because it's a constant
coefficient and we have the integral of the quantity ln of x to the n minus one dx and
as you can see we have up here in our reduction formula x ln of x to the n minus n times the
integral of ln of x to the n minus dx and that's exactly what we have here so that's
our we use integration by parts to prove this reduction formula and now if you get ever
this integral in exactly this form where you have ln of x raised to a constant instead
of going through the integration by parts process and finding u and dv and taking getting
du and v and plugging this all in and figuring out what the simplified integral is you could
just use this reduction formula to go directly to this step and skip the integration by parts
process so that's why we use reduction formulas and that's who you can use integration by
parts to find the reduction and prove this formula so I you found that video helpful
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